From 12f808b886d79607e5db7950525dc4804718f44d Mon Sep 17 00:00:00 2001
From: Zaran <zaran.krleza@gmail.com>
Date: Tue, 8 May 2012 01:17:06 -0700
Subject: Shapley value part

---
 slides.tex | 37 +++++++++++++++++++++++++++++++++----
 1 file changed, 33 insertions(+), 4 deletions(-)

(limited to 'slides.tex')

diff --git a/slides.tex b/slides.tex
index cd5547f..7d19984 100644
--- a/slides.tex
+++ b/slides.tex
@@ -8,8 +8,7 @@
 \usepackage{tikz}
 \usetheme{Boadilla}
 \usecolortheme{beaver}
-\title[Data monetization]{Data monetization: what is the value of your
-data?}
+\title[Data monetization]{Data monetization: pricing user data with the Shapley Value}
 \author{Thibaut \textsc{Horel}}
 \institute[Technicolor]{Work with {\greektext Στρατής \textsc{Ιωαννίδης}}}
 \setbeamercovered{transparent}
@@ -48,18 +47,48 @@ data?}
 \section{Shapley value}
 
 \begin{frame}{Individual value}
+  \begin{itemize}
+  \item $S$ the set of players, a subset $S'\subset S$ is a \alert{coalition}
+  \item $V: \mathcal{P}(S)\to \mathbf{R}$ the \alert{value} function:
+    \begin{itemize}
+    \item $V(S')$ is the value of the coalition $S'$
+    \item $V(\emptyset) = 0$
+    \end{itemize}
+  \end{itemize}
 
+  \alert{Question:} How to split the value of a subset across all its constituents?
+
+  \alert{Goal:} design a function $\phi_i(S',V)$, the value of player $i$ in $S'$
+
+  \begin{itemize}
+    \visible<2->{\item $\phi_i(S',V) = V(\{i\})$?} \visible<3->{``The whole is greater than the sum of its parts''
+      $$V(S')\neq \sum_{i\in S'} V(\{i\})$$}
+    \visible<4->{\item $\phi_i(S',V) = V(S') - V(S'\setminus\{i\})$: the \alert{marginal contribution}?} \visible<5->{depends on the time the player joins the coalition}
+  \end{itemize}
 \end{frame}
 
 \begin{frame}{Axioms}
+  \begin{description}
+    \item[Efficiency:] The whole value must be split
+      $$V(S) = \sum_{i\in S} \phi_i(S,V)$$
+    \item[Symmetry:] Equal pay for equal contribution
+
+      \alert{if} $\forall S'\subset S\setminus\{i,j\}, V(S'\cup\{i\}) = V(S'\cup\{j\})$
 
+      \alert{then} $\phi_i(S,V) = \phi_j(S,V)$
+    \item[Fairness:] $i$'s contribution to $j$ equals $j$'s contribution to $i$
+      $$\phi_i(S,V) - \phi_i(S\setminus\{j\}) = \phi_j(S,V)- \phi_j(S\setminus\{i\})$$
+  \end{description}
 \end{frame}
 
-\begin{frame}{The solution}
+\begin{frame}{Solution: the Shapley Value}
 
+  \begin{theorem}[Shapley, 1953]
+
+  \end{theorem}
 \end{frame}
 
-\begin{frame}{How to use the Shapley Value}
+\begin{frame}{Properties}
 
 \end{frame}
 
-- 
cgit v1.2.3-70-g09d2

